PART A )Rank from most concentrated to least concentrated. 1 )200. mL of 1.50 M

Question

PART A )Rank from most concentrated to least concentrated.

1 )200. mL of 1.50 M NaCl solution

2) 100. mL of 3.00 M NaCl solution

3) 150. mL of solution containing 21.0g of NaCl

PART B

A student placed 19.5g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 35.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution?

4) 100. mL of solution containing 21.0g of NaCl

5) 300. mL of solution containing 0.450 mol NaCl

Explanation / Answer

PART A:

1 )200. mL of 1.50 M NaCl solution

moles of NaCl=0.2*1.5=0.3 , molarity=1.5 M

2) 100. mL of 3.00 M NaCl solution

moles of NaCl=0.1*3.00=0.3, molarity=3.00 M

3) 150. mL of solution containing 21.0g of NaCl

moles of NaCl=21.0/58.5 =0.3589, molarity=0.3589/0.15 = 2.3931 M

4) 100. mL of solution containing 21.0g of NaCl

moles of NaCl=21/58.5 =0.3589 , molarity=0.3589/0.1 =3.589 M

5) 300. mL of solution containing 0.450 mol NaCl

molarity= moles/volume in litre =0.45/0.3= 1.5 M

So, order of cocnentration is as follows:
2>4>3>1=5

PART B

Moles of glucose in 19.5 g of it=19.5/180.15= 0.108

concentration of solution when volume of solution was 100 ml = moles*voluem in litre=0.108*0.1=0.0108 M

Further a 35.0 mL of this 0.0108 M glucose solution was diluted to 0.500 L.

so, moles of glucose in 35 ml of 0.0108 M solution=0.0108*0.035=0.000378

Now, as the final volume is made to 0.5 L, concentrtion of solution = moles of glucose present/volume

=0.000378/0.5= 0.000756 M

moles of this 0.000756 M glucose in 100 ml of solution = 0.000756*0.1= 0.0000756

weight in grams of glucose in 0.0000756 moles of it= 0.0000756*180.15= 0.13619 g

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