Chemistry · 2019
JEE Main · 12 April 2019, Shift 2 · Q55
A solution is prepared by dissolving 0.6 g of urea (molar mass =60 g mol^-1 ) and 1.8 g of glucose (. molar mass.=180 g mol^-1) in 100 mL of water at…
A solution is prepared by dissolving $\displaystyle 0.6$ g of urea (molar mass $\displaystyle =60 \mathrm{~g} \mathrm{~mol}^{-1}$ ) and $\displaystyle 1.8$ g of glucose $\displaystyle \left(\right.$ molar mass $\displaystyle \left.=180 \mathrm{~g} \mathrm{~mol}^{-1}\right)$ in $\displaystyle 100$ mL of water at $\displaystyle 27$ °C. The osmotic pressure of the solution is :
$$\left(\mathrm{R}=0.08206 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)
$$
Official answer
(1)
4.$\displaystyle 92$ atm
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JEE Main 2019 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.