If O 2 gas is bubbled through water at 303 K, the number of millimoles of O 2 gas that dissolve in 1 litre of water is (Nearest Integer) (Given: Henry's Law constant for O 2 at 303 K is 46.82 k bar and partial pressure of O 2= 0.920 bar) (Assume solubility of O 2 in water is too small, nearly negligible)

Question

If O 2 gas is bubbled through water at 303 K, the number of millimoles of O 2 gas that dissolve in 1 litre of water is (Nearest Integer) (Given: Henry's Law constant for O 2 at 303 K is 46.82 k bar and partial pressure of O 2= 0.920 bar) (Assume solubility of O 2 in water is too small, nearly negligible) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

p = K H × x underset textbar0.920=46.82 × 103 text bar × ( text mol of O 2/ text mol of H 2 O ) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/63f477fe73b5e3d5648f070c42e433e6-.png" /> 0.920=46.82 × n o 2 p =(0.920/46.82 × 18)= n 02 ⇒ 1.09 × 10-3= n 02 ⇒ m mol of 2=1

p=KH​×x bar0.920​=46.82×103 bar × mol of H2​O mol of O2​​ 0.920=46.82×no2​​ p=46.82×180.920​=n02​​ ⇒1.09×10−3=n02​​ ⇒mmolof2​=1

$p = K_{H} \times x$
$bar 0.920 = 46.82 \times 1 0^{3} bar \times \frac{mol of O _{2}}{mol of H _{2} O}$

$0.920 = 46.82 \times n_{o_{2}}$
$p = \frac{0.920}{46.82 \times 18} = n_{0_{2}}$
$\Rightarrow 1.09 \times 1 0^{- 3} = n_{0_{2}}$
$\Rightarrow m m o l o f_{2} = 1$