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Standard XII

Chemistry

Van't Hoff Factor

A 5.25 % solu...

Question

A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (Molar mass $= 60 g/mol$ ) in the same solvent.
If the densities of both the solutions are assumed to be equal to $1 {g/cm}^{3}$ , the molar mass of the substance will be:

210 g/mol

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90 g/mol

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115 g/mol

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105 g/mol

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Solution

The correct option is A $210 g/mol$
Since the solutions are isotonic, the concentration of the substance = concentration of urea
$∴ \frac{52.5 g/l}{M} = \frac{15 g/l}{60 g/mol}$

$(∵ 100 g solution = 100 ml as d = 1 g/ml$
and $1 L = 1000 ml)$

$M = \frac{52.5 \times 60}{15} = 210 g/mol$

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Similar questions

A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molecular mass = 60g/mole) in the same solvent . If the densities of both the solutions are assumed to be equal to 1g/cc, molar mass of substance will be

Q. A

5.25

solution of a substance is isotonic with a

1.5 %

solution of urea (molar mass =

60 g m o l^{- 1})

in the same solvent. If the densities of both the solutions are assumed to be equal to

1.0 g c m^{- 3}

, molar mass of the substance will be:

Q. A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol

^{- 1}

) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 gcm

^{- 3}

, molar mass of the substance will be :

Q. At certain temperature,

1.6

% solution of an unknown substance is isotonic with

2.4

% solution of urea. If both the solutions have the same solvent and both the solution have same density

1 g m / {c m}^{3}

, what will be the molecular mass of unknown substance in

g m / m o l

.
(Molecular mass of urea

60 g m / m o l

)

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A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (Molar mass =60 g/mol) in the same solvent. If the densities of both the solutions are assumed to be equal to 1 g/cm3, the molar mass of the substance will be:

The correct option is A 210 g/molSince the solutions are isotonic, the concentration of the substance = concentration of urea ∴52.5 g/lM=15 g/l60 g/mol (∵100 g solution=100 ml as d=1 g/ml and 1 L=1000 ml) M=52.5×6015=210 g/mol

A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (Molar mass $= 60 g/mol$ ) in the same solvent.
If the densities of both the solutions are assumed to be equal to $1 {g/cm}^{3}$ , the molar mass of the substance will be:

The correct option is A $210 g/mol$
Since the solutions are isotonic, the concentration of the substance = concentration of urea
$∴ \frac{52.5 g/l}{M} = \frac{15 g/l}{60 g/mol}$

$(∵ 100 g solution = 100 ml as d = 1 g/ml$
and $1 L = 1000 ml)$

$M = \frac{52.5 \times 60}{15} = 210 g/mol$