3) Calculate the molalities of the following aqueous solutions: a)1.22 Mglucose

Question

3) Calculate the molalities of the following aqueous solutions: a)1.22 Mglucose (C&H1n06;) solution (density of solution = 1 12 ginL), (b) 087 MNaOH solution (density of solution-104 g mL), (c) S 24 MNaHCOs solution (density of solution = 1.19 g/mL) 4) Specify what ions are present upon dissolving each of the following substances in water: (a) Mglb, (b) AINO), (c) HCI04, (d) NaCHsCOO 5) In the experiment of using freezing-point depression to find the molar mass of a compound, a solution of 3.80 g of a compound having the empirical formula CoHsP in 19.6 g of benzene is observed to freeze at 3.2°C. Calculate the molar mass of the solute and its molecular formula. (For benzene, the normal freezing point is 5.5°C and Ke value is 5.12°C/m)

Explanation / Answer

Q3.

calculate molality

Given

molarity =1.22 M means 1.22 mole of glucose present in 1 litre solution

density=1.12g/ml

density=mass/volume

so, 1.12=mass of solution/1000ml

mass of solution = 1120

mole of glucose are 1.22

mole = wt of solute/moleculer wt.

1.22=wt of solute/180

wt of solute = 219.60

so, wt of solvent = wt of soltion - wt of solute

=1120-219.60

=900.4 g

=0.9kg

molality= moles of solute/wt of solvent in kg

=1.22/0.9

=1.355

b)

Given

molarity =0.87 M means 0.87mole of NaOH present in 1 litre solution

density=1.04g/ml

density=mass/volume

so, 1.04=mass of solution/1000ml

mass of solution = 1040

mole of NaOH are 0.87

mole = wt of solute/moleculer wt.

0.87=wt of solute/40

wt of solute = 34.80

so, wt of solvent = wt of soltion - wt of solute

=1040-34.80

=1005.2 g

=1.0052kg

molality= moles of solute/wt of solvent in kg

=0.87/1.0052

=0.865

c)

Given

molarity =05.24 M means 5.24mole of NaHCO3 present in 1 litre solution

density=1.19g/ml

density=mass/volume

so, 1.19=mass of solution/1000ml

mass of solution = 1190 g

mole of NaHCO3=5.24

mole = wt of solute/moleculer wt.

5.24=wt of solute/84

wt of solute = 440.16 g

so, wt of solvent = wt of soltion - wt of solute

=1190-440.19

=749.84

=0.749kg

molality= moles of solute/wt of solvent in kg

=5.24/0.749

=6.99

Q4)

MgI2 = Mg2+ and I-

Al(NO3)3 = Al3+ and NO3-  

HClO4 = H+ and ClO4-

NaCH3COO is written as CH3COONa = CH3COO- and Na+

Q5.

lets moleculer wt will be X

wt of solute =3.8g

wt of solvent =19.6g of benzene

Tf= 3.2oC

Tof= 5.5oC

Delta Tf= Kf*m

m for molality=moles of solute/wt of solvent in kg

Tof - Tf= kf*(moles of sulte/wt of solvent in g)*1000

5.5-3.2=5.12*(3.8/X*19.6)*1000

X=431.58

we know that

n= molecular wt/emperical wt

so for C6H5p emperical wt =108

moleculaer wt =431.52

n =431.58/108

= 4(approx)

so the mlecular formula is 4C6H5p

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