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Question

The vapour pressure of acetone at $$20^{\circ}$$C is $$185$$ torr. When $$1.2$$ g of a non-volatile substance was dissolved in $$100$$ g of acetone at $$20^{\circ}$$C, its vapour pressure was $$183$$ torr. 
The molar mass ($$g mol$$$$^{-1}$$) of the substance is:


A
32
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B
64
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C
128
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D
488
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Solution

The correct option is B $$64$$
The molar mass $$M_2$$ of non-volatile substance can be calculated from the following formula.

$$\dfrac {p^0-p}{p^0} =\dfrac {W_2 \times M_1}{W_1 \times M_2} $$

Here, $$p^0$$ vapour pressure of acetone and $$p$$ is the vapour pressure of the solution.
$$M_1$$ and $$W_1$$ are the molar mass and mass of acetone.
$$M_2$$ and $$W_2$$ are the molar mass and mass of a non-volatile substance.

Substitute values in the above expression.

$$\dfrac {185-183}{185} =\dfrac {1.2 \times 58}{100 \times M_2} $$

Hence, $$M_2=64.38\: g/mol$$.
Thus, the molar mass of the non-volatile substance is $$64$$ $$g/mol$$ (approximately).
Hence, the correct option is $$\text{B}$$

Chemistry

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