Calculate the molarity of a solution having 3.42 g sucrose in 500 mL solution.

0.2 M

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0.02 M

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2 M

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0.342 M

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Solution

The correct option is B
0.02 M

We know that,
$M o l a r i t y = \frac{N u m b e r o f m o l e s o f s o l u t e}{N u m b e r o f l i t r e s o f s o l u t i o n}$

We know the given weight of sucrose from which the number of moles of sucrose can be calculated.

Therefore $N u m b e r o f m o l e s = \frac{g i v e n w e i g h t}{M o l e c u l a r w e i g h t}$
$= \frac{3.42}{342} = 0.01$

When 0.01 moles of sucrose is added to 500 mL

$\Rightarrow M o l a r i t y = 0.01 \times \frac{1}{500 \times 10^{- 3}} (1 m L = 10^{- 3} L)$

$\Rightarrow M o l a r i t y = 0.02 M$

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