Question
Asked Nov 12, 2019
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How many grams of solute are needed to make 5.00 × 102 mL of 0.445 M CH3OH?

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Expert Answer

Step 1

Given,

Molarity = 0.445 M

Volume = 5.00 x 102 mL

Conversion of volume, mL to L:
5 x 100 mL 500 mL
1000 mL 1 L
500
-mL= 0.5L
1000
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Conversion of volume, mL to L: 5 x 100 mL 500 mL 1000 mL 1 L 500 -mL= 0.5L 1000

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Step 2

The formula of mo...

No.of moles of solute
Molanity
Volume of solution in litres
No.of moles of solute= Molarity x Volume of solution in litres
0.445 mol
x0.5 L
= 0.2225 mol
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No.of moles of solute Molanity Volume of solution in litres No.of moles of solute= Molarity x Volume of solution in litres 0.445 mol x0.5 L = 0.2225 mol

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