What is the final concentration if 25.0 mL of 2.50 M glucose solution is diluted

Question

What is the final concentration if 25.0 mL of 2.50 M glucose solution is diluted to a volume of 550.0 mL ? Copper reacts with dilute nitric acid according to the following balance equation. If copper wire weighing 2.0 g is completely dissolved in a small amount of nitric acid and resultant solution is diluted to 150.0 mL of water, what is the molarity of Cu(NO_3) ? 3Cu (s) + 3HNO_2 (aq) rightarrow 3Cu (NO_3) + 2NO (g) + 4H_2 O (l) An unknown liquid is composed of 5.57 % H, 28.01 % Cl and 66.47% C. The molecular mass found by mass is 125.58 a.m.u. What is the molecular formula of the unknown compound?

Explanation / Answer

1 ) We can use the relation,

N1 V1 = N2 V2

Here , N1 = 2.50 M, N2 = ....

V1 = 25 mL, V2  = 550 mL

Therefore,

2.50 x 25 = N2 x 550

therefore, N2  = 2.50 x 25 / 550

= 0.11 M

Concentration of 550 ml diluted solution = 0.11 M

2 ) Problem is Not clearly visible in image.

3 ) % H + % Cl + % C = 5.57 + 28.01 + 66.42 = 100 %

The total percentage is 100 so there is no Oxygen in compound.

Then, No. of moles of H = % H / Atomic mass of Hydrogen = 5.57 / 1 = 5.57

No. of moles of Cl = % Cl / Atomic mass of Chlorine = 28.01 / 35.5 = 0.789

No. of moles of C = % C / Atomic mass of Carbon = 66.42 / 12 = 5.53  

The ratio of number of moles is, 5.57 / 0.789 = 7.0, 0.789 / 0.789 = 1 & 5.53 / 0.789 = 7 (Divided by lowest value of number of moles )

Hence empirical formula is H7Cl1C7 = C7H7Cl

Empirical formula wt = 84+7+35.5 126.5

ie Empirical wt ( 126.5 ) = Molecular wt ( 126.58 )

Therefore Moleculr formula = C7H7Cl

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