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Q.

The correct relationship between molarity (M) and molality (m) is (d = density of the solution, in Kg L^-1, M₂ = molar mass of the solute in kg mol^-1)

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a

large M, = ,frac{{md}}{{1 + m{M_2}}}

b

M, = ,frac{m}{{1 + m{M_2}d}}

c

M, = ,frac{{1 + m{M_2}}}{{md}}

d

M, = ,frac{{1 + md}}{{m{M_2}}}

answer is A.

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Detailed Solution

Given M=Molarity of solution ; m=molality of solution

M₂=MW of solute in kg ; d= density of solution in kg/lit

$molality = frac{no.of;moles;of;solute(n)}{1;kg;of;solvent}$

molality = n_solute (in magnitude)

W_solvent = 1 kg

W_solute(in kg) = n_solute X MW of solute in kg

we know , n_solute= m ; MW of solute in kg = M₂

$therefore ;W_{solute};=mM_2$

$W_{solution};=W_{solvent}+W_{solute};;(in;kg)=1+mM_2$

$d_{solution}=frac{W_{sol}in;kg}{V_{sol}inlit}$

$V_{sol}=frac{W_{sol}}{d_{sol}}=frac{1+mM_2}{d}$

$Molarity (M)= n_{solute}Xfrac{1}{V_{solution}}$

$=m;X;frac{1}{frac{1+mM_2}{d}}$

$therefore;M=frac{md}{1+mM_2}$

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