Imat 2021 chemistry q43

Practice Question Solving
1

Hi can someone kindly solve this please I’m not sure how to work out the dilution factor

100 mL of a solution contained 3.2 g of potassium nitrate

25 mL of this solution was added to an empty 250 mL volumetric flask. Distilled water was added up to the 250 mL mark and the flask was shaken to ensure that mixing was complete.

A pipette was used to transfer 25 mL of the resulting solution from the volumetric flask to an empty conical flask.

–1 What is the concentration of the potassium nitrate solution, in g L, in the conical flask?

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2

Formula: C1*V1=C2V2
Initial conc. *initial vol=final conc.final vol
To calculate C1 (gL-1)=Mass/vol = 3.2/100(10-3) =3.2/0.1 =32 gl-1
V1=25mL=0.025L
C2=? (we have to calculate)
V2=250ml=0.25L
Putting values in formula;
32×0.025=C20.25
C2=32×0.025/0.25
C2=32×0.1=3.2gL-1

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3.2g KNO3 in 100Ml of solution (given)

thus, in 25ml we would have 0.8g KNO3

Upon adding water till 250Ml we essentially increase the volume of the solution to 250Ml.

thus we have 0.8g in 250Ml out of which we take 25Ml which amounts to 0.08g in 25Ml

Therefore in the end the concentration in g/L= (0.08/25)*1000 = 3.2g/l

alternatively c1v1=c2v2 can also be used like mentioned by @Ibraheem_Saeed

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Ahh thank you, you’ve both explained it really clearly.

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