Sea water is found to contain 5.85 % NaCl and 9.50 % MgCl 2 be weight of solution. Calculate its normal boiling point assuming 80 % ionisation for NaCl and 50 % ionisation of MgCl 2[ K b( H 2 O ). =0.51 kg mole -1 K ].

Question

Sea water is found to contain 5.85 % NaCl and 9.50 % MgCl 2 be weight of solution. Calculate its normal boiling point assuming 80 % ionisation for NaCl and 50 % ionisation of MgCl 2[ K b( H 2 O ). =0.51 kg mole -1 K ]. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

100 gm longrightarrow 5.85 gm NaCl =.01 mole NaCl 100 gm longrightarrow 9.5 gm MgCl 2=.1 mole MgCl 2 wt. of water =100-5.85-9.5=84.65 gm Δ T b =(0.51) ((5.85/58.5)(1+0.82)+(9.5/95)(1+0.5)/84.651) T b -100=(0.51)[(0.18+0.15/84.651))(1000) =((0.51)(0.33)(1000)/84.651) T b =100+1.988 T b =101.988° C

Q. Sea water is found to contain 5.85%NaCl and 9.50%MgCl2​ be weight of solution. Calculate its normal boiling point assuming 80% ionisation for NaCl and 50% ionisation of MgCl2​[Kb​(H2​O)=0.51kgmole−1K].

100gm⟶5.85gmNaCl=.01 mole NaCl 100gm⟶9.5gmMgCl2​=.1 mole MgCl2​ wt. of water =100−5.85−9.5=84.65gm ΔTb​=(0.51)84.65158.55.85​(1+0.82)+959.5​(1+0.5)​ Tb​−100=(0.51)[84.6510.18+0.15​)(1000) =84.651(0.51)(0.33)(1000)​ Tb​=100+1.988 Tb​=101.988∘C

Q. Sea water is found to contain $5.85% N a Cl$ and $9.50% M g C l_{2}$ be weight of solution. Calculate its normal boiling point assuming $80%$ ionisation for $N a Cl$ and $50%$ ionisation of $M g C l_{2} [K_{b} (H_{2} O) = 0.51 k g m o l e^{- 1} K]$ .