When $W_B \ g$ of solute (molecular mass $M_B$) is dissolved in $W_A \ g$ of solvent,the molality $m$ of the solution is:
- A$\frac{W_B}{W_A} \times \frac{M_B}{1000}$
- B$\frac{W_B}{M_B} \times \frac{1000}{W_A}$
- C$\frac{W_A}{W_B} \times \frac{1000}{M_B}$
- D$\frac{W_A \times M_B}{W_B \times 1000}$
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Commercially available $H_2SO_4$ is $98 \ g$ by weight of $H_2SO_4$ and $2 \ g$ by weight of water. Its density is $1.38 \ g \ cm^{-3}$. Calculate the molality $(m)$ of $H_2SO_4$ (molar mass of $H_2SO_4$ is $98 \ g \ mol^{-1}$). (in $m$)
$6 \ g$ of urea (molar mass $= 60 \ g \ mol^{-1}$) and $9 \ g$ of glucose (molar mass $= 180 \ g \ mol^{-1}$) were dissolved in $35 \ g$ of water. The mass percent of urea and glucose is respectively:
The molality of a solution,when $18 \ g$ of glucose is added to $18 \ g$ of $H_2O$,is: (in $m$)
What is meant by the concentration of a solution?
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View SolutionThe molarity of the solution when $4.9 \ g$ of $H_2SO_4$ is dissolved in $250 \ mL$ of solution is: (in $M$)
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