Solution A contains 7g/L of MgCl2 and Solution B contains 7g/L of NaCl. At room temp, Osmotic pressure ofa)A Bb)A = Bc)B A d)Insufficient informationCorrect answer is option 'C'. Can you explain this answer?

Question

Solution A contains 7g/L of MgCl2 and Solution B contains 7g/L of NaCl. At room temp, Osmotic pressure ofa)A > Bb)A = Bc)B >A   d)Insufficient informationCorrect answer is option 'C'. Can you explain this answer?

Answer

To calculate the osmotic pressure of a solution, we can use the formula:

Osmotic Pressure = (n/V)RT

Where:
n = number of moles of solute
V = volume of solution in liters
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin

In this case, we are given that Solution A contains 7g/L of MgCl2. To find the number of moles of MgCl2, we need to divide the mass by the molar mass.

The molar mass of MgCl2 is:
(1 x atomic mass of Mg) + (2 x atomic mass of Cl)
= (1 x 24.31 g/mol) + (2 x 35.45 g/mol)
= 24.31 g/mol + 70.90 g/mol
= 95.21 g/mol

Now, we can calculate the number of moles of MgCl2 in Solution A:
Number of moles = (mass of MgCl2) / (molar mass of MgCl2)
= 7g / 95.21 g/mol
= 0.0735 mol

Since Solution A contains 7g/L of MgCl2, we can consider the volume of the solution to be 1 liter (1000 mL).

Now, we can calculate the osmotic pressure of Solution A:
Osmotic Pressure of A = (n/V)RT
= (0.0735 mol / 1 L) * (0.0821 L·atm/(mol·K)) * (298 K)
= 2.18 atm

Therefore, the osmotic pressure of Solution A is 2.18 atm.

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