For $$0.6g$$ of a solute is dissolved in $$0.1$$ litre of a solvent which develops an osmotic pressure of $$1.23atm$$ at $${27^o}C$$ . The molecular mass of the substance is?
(A) $149.5gmo{l^{ - 1}}$
(B) $120gmo{l^{ - 1}}$
(C) $430gmo{l^{ - 1}}$
(D) $102gmo{l^{ - 1}}$
Answer
539.4k+ views
Hint: Colligative properties can be defined as the properties of a liquid which depend upon the amount (number of moles) of the solute that is present in that solution (liquid). Osmotic pressure can be defined as the pressure that must be applied to the solution side to halt or stop the flow of solvent particles or molecules through a semipermeable membrane (osmosis).
Complete answer:
We know that colligative properties can be defined as the properties of a liquid which depend upon the amount (number of moles) of the solute that is present in that solution (liquid).
We can define osmotic pressure as the pressure that must be applied to the solution side to halt or stop the flow of solvent particles or molecules through a semipermeable membrane (osmosis).
Formula for osmotic pressure is:
$\pi = C \times R \times T$
Where, $\pi $ is the osmotic pressure, C is the concentration of solute, T is the temperature and R is the gas constant.
It can be written as:
$\pi = \dfrac{{m \times R \times T}}{{M \times V}}$
So, molar mass of the solute can be calculated as:
$M = \dfrac{{m \times R \times T}}{{\pi \times V}}$
$$R = 0.0821Latm{K^{ - 1}}mo{l^{ - 1}}$$
$$\pi = 1.23atm$$
$$T = 300K$$
Now put these values in the formula for molar mass:
$M = \dfrac{{0.6 \times 0.0821 \times 300}}{{1.23 \times 0.1}}$
$M = 120gmo{l^{ - 1}}$
The molar mass of the solute is found to be $120gmo{l^{ - 1}}.$
Hence, the correct option is (B) $120gmo{l^{ - 1}}.$
Note:
Now we know that colligative properties can be defined as the properties of a liquid which depend upon the amount (number of moles) of the solute that is present in that solution (liquid). There are four colligative properties such as vapour pressure lowering, depression in freezing point, elevation in boiling point and osmotic pressure.
Complete answer:
We know that colligative properties can be defined as the properties of a liquid which depend upon the amount (number of moles) of the solute that is present in that solution (liquid).
We can define osmotic pressure as the pressure that must be applied to the solution side to halt or stop the flow of solvent particles or molecules through a semipermeable membrane (osmosis).
Formula for osmotic pressure is:
$\pi = C \times R \times T$
Where, $\pi $ is the osmotic pressure, C is the concentration of solute, T is the temperature and R is the gas constant.
It can be written as:
$\pi = \dfrac{{m \times R \times T}}{{M \times V}}$
So, molar mass of the solute can be calculated as:
$M = \dfrac{{m \times R \times T}}{{\pi \times V}}$
$$R = 0.0821Latm{K^{ - 1}}mo{l^{ - 1}}$$
$$\pi = 1.23atm$$
$$T = 300K$$
Now put these values in the formula for molar mass:
$M = \dfrac{{0.6 \times 0.0821 \times 300}}{{1.23 \times 0.1}}$
$M = 120gmo{l^{ - 1}}$
The molar mass of the solute is found to be $120gmo{l^{ - 1}}.$
Hence, the correct option is (B) $120gmo{l^{ - 1}}.$
Note:
Now we know that colligative properties can be defined as the properties of a liquid which depend upon the amount (number of moles) of the solute that is present in that solution (liquid). There are four colligative properties such as vapour pressure lowering, depression in freezing point, elevation in boiling point and osmotic pressure.
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