Question:
Determine the amount of \(CaCl_2 (i = 2.47)\) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at \(27\degree C\).
Determine the amount of \(CaCl_2 (i = 2.47)\) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at \(27\degree C\).
Updated On: June 02, 2025
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Solution and Explanation
The correct answer is: 3.42 g
We know that,
\(π=i\frac{n}{V}RT\)⇒π=iw/MVRT
\(⇒w=\frac{πMV}{iRT}\)
\(π=0.75atm\)
V=2.5L
i=2.47
T=(27+273)K=300K
Here,
\(R = 0.0821\, L\, atm\, K^{-1} mol^{-1 }\)
\(M = 1 \times 40 + 2 \times 35.5 \)
\(= 111g mol^{-1}\)
Therefore, \(w=\frac{0.75 \times 111 \times 2.5}{2.47 \times 0.0821 \times 300}\)
= 3.42 g
Hence, the required amount of \(CaCl_2\) is 3.42 g.
We know that,
\(π=i\frac{n}{V}RT\)⇒π=iw/MVRT
\(⇒w=\frac{πMV}{iRT}\)
\(π=0.75atm\)
V=2.5L
i=2.47
T=(27+273)K=300K
Here,
\(R = 0.0821\, L\, atm\, K^{-1} mol^{-1 }\)
\(M = 1 \times 40 + 2 \times 35.5 \)
\(= 111g mol^{-1}\)
Therefore, \(w=\frac{0.75 \times 111 \times 2.5}{2.47 \times 0.0821 \times 300}\)
= 3.42 g
Hence, the required amount of \(CaCl_2\) is 3.42 g.
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CBSE CLASS XII Notification
Concepts Used:
Osmosis
Osmosis is a passive process and happens without any expenditure of energy. It involves the movement of molecules from a region of higher concentration to lower concentration until the concentrations become equal on either side of the membrane.
There are three different types of solutions:
- Isotonic Solution is one that has the same concentration of solutes both inside and outside the cell.
- Hypertonic solution is one that has a higher solute concentration outside the cell than inside.
- Hypotonic solution is one that has a higher solute concentration inside the cell than outside.