Question

A 5% solution of cane sugar (molar mass 342) is isotonic
with 1% of a solution of an unknown solute. The molar mass
of unknown solute in g/mol is :
(a) 171.2 (b) 68.4 (c) 34.2 (d) 136.2

Answer 1

The molar mass of unknown solute in a 5% of solution of cane sugar with 1% of a solution of an unknown solute is 68.4

• Given,
• Molar concentration of solution,
• $x=\dfrac{1}{m} * \dfrac{1000}{100}=\dfrac{10}{m}$....(1)
• But,
• molar concentration of cane sugar,
• $=\dfrac{5}{342} * \dfrac{1000}{100}=\dfrac{50}{342}$.....(2)
• equating equations (1) and (2), we get,
• $\dfrac{10}{m}=\dfrac{50}{342}$
• ⇒$m=68.4$

Answer 2

The molar mass of the solute is m = 68.4  

Option (B) is correct.

Explanation:

Molar concentration of cane sugar:

= 5 / 342 x 1000 / 100 = 50 / 342

Molar concentration of solution.

X = 1/m x 1000 / 100 x= 10 /m

10 / m = 50 / 342

m x 50 = 342 x 10

m = 342 x 10 / 50

m = 68.4

Thus the molar mass of the solute is m = 68.4  

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