36.8 g of CuSO4(s) was added to water to prepare a 2.00 L solution. What is the
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We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.
The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³
The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.
Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.
5.71 grams of Na₂CO₃.10 H₂O is equal to = 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.
Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is = 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³
Answer : The partial pressure of and
is, 216.5 mmHg and 649.5 mmHg
Explanation :
According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.
Formula used :
So,
where,
= partial pressure of gas
= mole fraction of gas
= total pressure of gas
= moles of gas
= total moles of gas
The balanced decomposition of ammonia reaction will be:
Now we have to determine the partial pressure of and
Given:
and,
Given:
Thus, the partial pressure of and
is, 216.5 mmHg and 649.5 mmHg