
The mole fraction of a solution is \[0\cdot \text{1}\] at 298 K molarity of this solution is the same as its molarity density of this solution at 298 K is 2.0 $gc{{m}^{3}}$. The ratio of the molecular weights of the solute and solvent MW solute MW solvent is?
Answer
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Hint: Molarity of a solute is given by the ratio of mass of solute and the molecular weight of the solute per litre. Mole fraction of a solute is the ratio of number of moles of solute to the number of moles of solution. Moles can be found by dividing the given mass by molecular mass.
Complete step by step solution:
Fraction of solute= $0\cdot 1$
Let, molecular weight of solute = m
Molecular weight of solvent = M
Let the volume of solution be 1000ml
Mass of solution will be 2000 g
Mass of solvent be x
Mass of solute will be 2000-x g
Molarity$=\dfrac{\dfrac{x}{m}}{litre}$
Molarity\[=\dfrac{\dfrac{x}{m}}{2000-x}\times {{10}^{-3}}\]
Equating, both the above equation, we get,
$\dfrac{x}{m}=\dfrac{1000x}{(2000-x)m}$
On simplifying, we get,
X= 1000 g
Number of moles of solute = $\dfrac{1000}{m}$
Number of moles of solvent = \[\dfrac{1000}{M}\]
$\begin{align}
& \dfrac{\dfrac{1000}{m}}{\dfrac{1000}{m}+\dfrac{1000}{M}}=\dfrac{M}{m+M}=0\cdot 1 \\
& M=0\cdot 1m+0\cdot 1M \\
& 0.9M=0\cdot 1m \\
& \dfrac{m}{M}=\dfrac{0\cdot 9}{0\cdot 1}=9 \\
\end{align}$
The ratio of molecular weight of solute and solvent is 9.
Note: The concept of molarity and molality should not be confused with each other. The moles of solute and solvent may not be the same always. Molarity is helpful in calculating the concentration of the solution. There also exists molality which determines the moles of solute with the mass of the solvent. Molarity is written as M and molality as m.
Complete step by step solution:
Fraction of solute= $0\cdot 1$
Let, molecular weight of solute = m
Molecular weight of solvent = M
Let the volume of solution be 1000ml
Mass of solution will be 2000 g
Mass of solvent be x
Mass of solute will be 2000-x g
Molarity$=\dfrac{\dfrac{x}{m}}{litre}$
Molarity\[=\dfrac{\dfrac{x}{m}}{2000-x}\times {{10}^{-3}}\]
Equating, both the above equation, we get,
$\dfrac{x}{m}=\dfrac{1000x}{(2000-x)m}$
On simplifying, we get,
X= 1000 g
Number of moles of solute = $\dfrac{1000}{m}$
Number of moles of solvent = \[\dfrac{1000}{M}\]
$\begin{align}
& \dfrac{\dfrac{1000}{m}}{\dfrac{1000}{m}+\dfrac{1000}{M}}=\dfrac{M}{m+M}=0\cdot 1 \\
& M=0\cdot 1m+0\cdot 1M \\
& 0.9M=0\cdot 1m \\
& \dfrac{m}{M}=\dfrac{0\cdot 9}{0\cdot 1}=9 \\
\end{align}$
The ratio of molecular weight of solute and solvent is 9.
Note: The concept of molarity and molality should not be confused with each other. The moles of solute and solvent may not be the same always. Molarity is helpful in calculating the concentration of the solution. There also exists molality which determines the moles of solute with the mass of the solvent. Molarity is written as M and molality as m.
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