A`5%` solution of cane sugar (molar mass =`342`)is isotonic with `1%`of a solution of an known solute.The molar mass of unknown solute in g/mol is

Options

A. `136.2`
B. `171.2`
C. `68.4`
D. `34.2`

Answer

C

Explanation

Two solution are said to be isotonic ,they have the same molar concentration,Therefore .
Concentration of cane sugar
=Concentration of `[C_(12)H_(22)O_(11)]`
As unknown solute`[X]`
for `C_(12)H_(22)O_(11)`,
concentration `=(5g)/(100cm^(3))=50//L=(50)/(342)molL^(-1)`
`[:. "Molar mass of" C_(12)H_(22)O_(11)=342g mol^(-1)]` for unknown substance X,
concentration ` =(1g)/(100cm^(3))=10g//L=(10)/(M)molL^(-1)`
Hence, `(50)/(342)molL^(-1)=(10)/(M)molL^(-1)`
`rArr M=(342xx10)/(50)`
`rArrM=68.4g molL^(-1)`