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Question

The molality of $$15 \% (wt./vol.)$$ solution of $$H_{2}SO_{4}$$ of density $$1.1 g/cm^{3}$$ is approximately.


A
1.2
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B
1.4
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C
1.8
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D
1.6
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Solution

The correct option is D $$1.6$$
The solution have 15% (mass/volume)

let mass of solute = 15g

volume of solution = 100 ml

we have, density = $$1.1 g/cm^3$$

so, mass of solution = $$density*volume = 1.1\times 100 = 110g$$
$$\text {mass of solvent(g)=} 110-15 g$$

molecular mass of $$H_2SO_4= 98$$ 

so, moles of $$H_2SO_4$$ = $$\dfrac{\text {given weight}}{\text {molecular weight}} = \dfrac{15}{98} = 0.153 moles$$ 

molality  = $$\dfrac{\text {moles of solute}}{\text {mass of solvent(g)}}\times 1000$$

so, molality =$$\dfrac{0.153}{95} \times 1000 = 1.6$$


Chemistry

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