Question

A solution containing 0.1 g of non-volatile organic substance p (molecular mass 100) in 100 g of benzene raises the boiling point of benzene by 0.2oc,0.2oc, while a solution containing 0.1g of another non-volatile substance q in the same amount of benzene raises the boiling point of benzene by 0.4oc0.4oc.What is the ratio of molecular masses of p and q ?

Answer 1

HELLO BHAISHAAB ,

The ratio is 2:1

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Answer 2

Answer:

The required ratio of the molecular masses is found to be 2:1.

Explanation:

We know that the expression for finding the boiling point for any substance is given by:

$\triangle T = K_bm_{solute}$

Where $m_{solute}$ is the molality of the solute (mass of solute in one kilo of solvent) which is given by the expression:

$m=\frac{w_B\times 1000}{m_B\times w_A}$

where $m$ represents the molecular mass and $w$ is the weight of solvent.

Now, here we can find the boiling point of substance P as:

$0.2 = K_b(\frac{0.1\times1000}{100\times 100})$             ...(i)

Now, letting the molecular mass of Q as $x$ and using the same expression above for substance Q, we get:

$0.4=K_b(\frac{0.1\times1000}{100\times x})$            ...(ii)

Now, dividing (i) by (ii), we get:

$\frac{0.2}{0.4} = \frac{K_b(\frac{0.1\times1000}{100\times 100})}{K_b(\frac{0.1\times1000}{100\times x})}$

Simplifying it, we get:

$\frac{1}{2} = \frac{x}{100}=50$

So, now we know the molecular masses of both substances P and Q as 100 and 50 respectively.

Thus, finding the ratio, we get:

$\frac{m_{BP}}{m_{BQ}}=\frac{100}{50}$

or we can say:

${m_{BP}}:{m_{BQ}}=2:1$

Thus, the required ratio is found to be 2:1

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