11th Class Chemistry Chapter 9 Solutions Short Questions Answers
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XA = nA XB = nB/nA + nB + nC, Xc = nc/ nA + nB + nC
nA+nB + nc ,
Adding these mole fractions:
XA + XB + Xc = nA/nA + nB+ nC + nB/ nA + nB + nC + nC / nA + nB + nc = 1
Molarity is the number of moles of the solute per dm3 of the solution. It has to the following formul:
Molarity (M) = Mass of solute 1
Molar mass of solute * Volume of solute in dm3
Molarity (M) = Number of moles of solu
Volume of solution in dm3
Molality (M)= Mass of solute
Molar mass of solute * Mass of solvent in kg
98% H2SO4 means that 100 g of H2SO4 solution has 98 g of H2 So4 and only 2 g of water.
Density = mass /volume
Since, volume = mass/ density
Putting the values
Volume = 100 g = 54.34cm3
1.84 g cm3
It means that the 98% H2SO4 having total mass of 100 g has a volume of 54.34.
Δp / P0 = X2
Vapour pressure and lowering of vaour pressure upon temperature. So, when the temperature of a solution is increased both the factors ΔP and P0 increase in such a way that the ratio remains the same.
Δp/p0 = X2
Since X2 = n2/n1 + n2 = n2/n1 (n1>> n2)
So, Δp/p0 = W2/M2 * M1/W1
or M2 = W2/W1 . M1 . p0/Δp
From these equation molar mass can be calculated.
MgCl2. 6H2O. AlCl3. 6H2O, Na2CO3, 10H2O, CuSO4, 5H2O
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