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pushpdeep
Answers (2)
molality= (Xb * 1000)/(1-Xb)*Ma Xb= mole fraction of nacl=0.5 Ma=18gm m=0.5*1000/0.5*18 m=55.56
44gms of CO2 is produced by (40+12+48)100gms of CaCO3 0.88gm of CO2 must be produced by (100/44)×0.88= 2gm of CaCO3 So the amount of pure CaCO3 present in 4gm of impure sample is 2gms. So percentage of purity = (2/4) ×100= 50
