Answer to Question #101942 in General Chemistry for jude

Question

2. a) Convert the following volumes into dm3:
i) 25 cm3
ii) 1300cm3

b) Calculate the concentrations of NaOH, in moldm-3, in the following solutions:
i) 2.0 moles of NaOH in 1dm3 of solution
ii) 1.0 mole of NaOH in 500cm3 of solution
iii) 0.05 moles of NaOH in 250cm3 of solution

c) Calculate the number of moles of solute in the following solutions, give your answers to 2 sig figs:

i) 10cm3 of a 1.00 moldm-3 solution
ii) 250cm3 of a 0.50 moldm-3 solution
iii) 2.0dm3 of a 0.20 moldm-3 solution
iv) 40cm3 of a 4.0 moldm-3 solution
v) 74cm3 of a 0.10 moldm-3 solution

d) Calculate the concentration, in moldm-3, of the following solutions:

i) 10g of NaCl in 1dm3 of solution
ii) 100g of CaBr2 in 200cm3 of solution
iii) 45g of CuSO4 in 3dm3 of solution
iv) 0.15g of NaOH in 25cm3 of solution
v) 5x10-2g of NaI in 100cm3 of solution


e) What mass of sodium carbonate (Na2CO3) would be needed to make 50cm3 of a 0.50moldm-3 solution?

Answer 1

a) $1dm^3=1000cm^3$

i) $25cm^3=0.025dm^3$

ii)$1300cm^3=1.3dm^3$

b) NaOH has a molecular weight of 40gms. $\implies 1 mol (NaOH)=40gms.$

i) $2 moldm^{-3}$

ii)$1/0.5=2moldm^{-3}$

iii)$0.05/0.25=0.2moldm^{-3}$

c) No. of moles = $Concentration*Volume$

i) $0.01*1=0.01 mol.$

ii) $0.25*0.5=0.125 \approx 0.12 mol.$

iii)$2*0.2=0.4 mol.$

iv)$0.04*4=0.16 mol.$

v)$0.074*0.1=0.0074=7.4 milimoles \approx 0.01mol.$

d) $Concentration=Moles/Volume$ $=Given weight/(Molecularweight*Volume)$

i) $10/(58.5*1)=0.171moldm^{-3}$

ii) $100/(120*0.2)=4.167moldm^{-3}$

iii) $45/(159.5*2)=0.141moldm^{-3}$

iv) $0.15/(40*0.025)=0.15moldm^{-3}$

v) $0.05/(150*0.1)=3.33*10^{-3}moldm^{-3}$

e) $No. of moles = Concentration*Volume$

$\implies 0.05*0.5=0.025mol=0.025*106=2.65gm.$