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Percentage Composition

How many mole...

Question

How many moles of $N H_{3}$ are there in $250 c m^{3}$ of a 30% solution, the specific gravity of which is 0.90?

3.97 moles

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6.37 moles

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3.70 moles

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3.50 moles

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Solution

The correct option is B 6.37 moles
Given value of this solution
volume of water is = 250 ml
density of solution is = 0.9 g/mL
Since ammonia dissolved in water is 30% by weight, then using mass % formula, we can calculate the amount of ammonia in water.
mass % = $\frac{m a s s o f a m m o n i a}{m a s s o f s o l u t i o n} \times 100$
30 = $\frac{m a s s o f a m m o n i a}{m a s s o f a m m o n i a} + m a s s o f w a t e r \times 100$
At 4oC, density of water = 0.999 g/mL
since density = $\frac{m a s s}{v o l u m e}$
0.999 = $\frac{m a s s}{250}$
mass of water = 249.75 g
now, let mass of ammonia be x.
$\frac{30}{100}$ = $\frac{x}{x + 249.75}$
30x +7492.5 = 100x
70x = 7492.5
x=107.03g
Thus, ammonia dissolved in 250mL water is 107.03 g

Moles of ammonia = mass of ammonia/molar mass
= $\frac{107.03 g}{17 g m o l^{- 1}}$

=6.3 moles
ans is B

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