(a)
Interpretation:
The moles and grams of the indicated solute in the given solution are to be calculated.
Concept Introduction:
The
The number of moles is calculated by the formula,
The molarity is calculated by the formula,
Answer to Problem 118AP
The moles and grams of the indicated solute in the given solution are
Explanation of Solution
The volume and molarity of
The molar mass of
The number of moles of a solute is calculated by the formula,
Substitute the values of volume of solution and molarity of
The mass of
Substitute the values of molar mass and number of moles of
Therefore, the moles and grams of the indicated solute in the given solution are
(b)
Interpretation:
The moles and grams of the indicated solute in the given solution are to be calculated.
Concept Introduction:
The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.
The number of moles is calculated by the formula,
The molarity is calculated by the formula,
Answer to Problem 118AP
The moles and grams of the indicated solute in the given solution are
Explanation of Solution
The volume and molarity of
The conversion of units of
The molar mass of
The number of moles of a solute is calculated by the formula,
Substitute the values of volume of solution and molarity of
The mass of
Substitute the values of molar mass and number of moles of
Therefore, the moles and grams of the indicated solute in the given solution are
(c)
Interpretation:
The moles and grams of the indicated solute in the given solution are to be calculated.
Concept Introduction:
The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.
The number of moles is calculated by the formula,
The molarity is calculated by the formula,
Answer to Problem 118AP
The moles and grams of the indicated solute in the given solution are
Explanation of Solution
The volume and molarity of
The conversion of units of
The molar mass of
The number of moles of a solute is calculated by the formula,
Substitute the values of volume of solution and molarity of
The mass of
Substitute the values of molar mass and number of moles of
Therefore, the moles and grams of the indicated solute in the given solution are
(d)
Interpretation:
The moles and grams of the indicated solute in the given solution are to be calculated.
Concept Introduction:
The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.
The number of moles is calculated by the formula,
The molarity is calculated by the formula,
Answer to Problem 118AP
The moles and grams of the indicated solute in the given solution are
Explanation of Solution
The volume and molarity of
The conversion of units of
The molar mass of
The number of moles of a solute is calculated by the formula,
Substitute the values of volume of solution and molarity of
The mass of
Substitute the values of molar mass and number of moles of
Therefore, the moles and grams of the indicated solute in the given solution are
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Chapter 15 Solutions
INTRODUCTORY CHEMISTRY
- 3.64 How many grams of solute are present in each of these solutions? (a) 37.2 mL ofO.471 M HBr (b) 113.0 L of 1.43 M Na2CO3 (c) 212 mL of 6.8 M CH3COOH (d) 1.3 × 10-4 L of 1.03 M H2S03arrow_forward34. For each of the following solutions, the number of moles of solute is given, followed by the total volume of the solution prepared. Calculate the molarity of each solution. a. 0.754 mol KNO; 225 mL b. 0.0105 in of CaCl; 10.2 mL c. 3.15 mol NaCl; 5.00 L d. 0.499 mol NaBr; 100. mLarrow_forward3.63 How many moles of solute are present in each of these solutions? (a) 48.0 mL of 3.4 M H2SO4. (b) 1.43 mL of 5.8 M KNO3. (c) 321 L of 0.034M NH3 (d) 1.9 × 10-3 L of 1.4 × 10-5 M NaFarrow_forward
- What is the difference between a solute and a solvent?arrow_forwardInsulin is a hormone that controls the use of glucose in the body. How many moles of insulin are required to make up 28 mL of 0.0048 M insulin solution?arrow_forwardA large beaker contains 1.50 L of a 2.00 M iron(III) chloride solution. How many moles of iron ions are in the solution? How many moles of chloride ions are in the solution? You now add 0.500 L of a 4.00 M lead(II) nitrate solution to the beaker. Determine the mass of solid product formed (in grams).arrow_forward
- When a solution is diluted by adding additional solvent, the concentration of solute changes hut the amount of solute present does not change. Explain. Suppose 250. mL of water is added to 125 mL of 0.55 1 M NaCl solution. Explain how you would calculate the concentration of the solution after dilution.arrow_forward3.61 Calculate the molarity of each of the following solutions. (a) 1.45 mol HCl in 250. mL of solution (b) 14.3 mol NaOH in 3.4 L of solution (c) 0.341 mol KCl in 100.0 mL of solution (d) 250 mol NaNO3 in 350 L of solutionarrow_forward3.65 Determine the final molarity for the following dilutions. (a) 24.5 mL of 3.0 M solution diluted to 100.0 mL (b) 15.3 mL of 4.22 M solution diluted to 1.00 L (c) 1.45 mL of 0.034 M solution diluted to 10.0 mL (d) 2.35 L of 12.5 M solution diluted to 100.0 Larrow_forward
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