A 1 % aqueous solution (M/V) of a certain substance is isotonic with a 3 % solution of glucose (M. Mass 180). The molar mass of the substance is

Question

A 1 % aqueous solution (M/V) of a certain substance is isotonic with a 3 % solution of glucose (M. Mass 180). The molar mass of the substance is (A) 60 (B) 120 (C) 180 (D) 360

Tardigrade Admin · Accepted Answer

For isotonic solutions . (n1/V1) = (n2/V2) (m1/M1 × V1) = (m2/M2 × V2) (1/M1 × 100) = (3/180 × 100) M1 = 60

Q. A 1 % aqueous solution (M/V) of a certain substance is isotonic with a 3 % solution of glucose (M. Mass 180). The molar mass of the substance is

60

120

180

360

For isotonic solutions .
$\frac{n _{1}}{V _{1}} = \frac{n _{2}}{V _{2}}$
$\frac{m _{1}}{M _{1} \times V _{1}} = \frac{m _{2}}{M _{2} \times V _{2}}$
$\frac{1}{M _{1} \times 100} = \frac{3}{180 \times 100}$
$M_{1} = 60$

Q. A 1 % aqueous solution (M/V) of a certain substance is isotonic with a 3 % solution of glucose (M. Mass 180). The molar mass of the substance is

60

120

180

360

For isotonic solutions . V1​n1​​=V2​n2​​ M1​×V1​m1​​=M2​×V2​m2​​ M1​×1001​=180×1003​ M1​=60

For isotonic solutions .
$\frac{n _{1}}{V _{1}} = \frac{n _{2}}{V _{2}}$
$\frac{m _{1}}{M _{1} \times V _{1}} = \frac{m _{2}}{M _{2} \times V _{2}}$
$\frac{1}{M _{1} \times 100} = \frac{3}{180 \times 100}$
$M_{1} = 60$