What is the molarity of an Nal solution that contains 6.0 g of Nal in 38.0 mL of

Question

What is the molarity of an Nal solution that contains 6.0 g of Nal in 38.0 mL of solution? 0.0055 M 0.00025 M 0.046 M 0.18 M 1.2 M A 27.0 g sample of NaOH in dissolved in water, und the solution is diluted to give a final volume of 1.10 L. The molarity of the final solution is 0.675 M. 0.0407 M. 24.5 M. 1.35 M. 0.614 M The concentration of sulfate in a sample of wastewater is to be determined by using gravimetric analysis. To a 100.0-mL sample of the wastewater is added an excess of lead nitrate, forming the insoluble lead sulfate (303.3 g/mol) according to the balanced equation given below. The solid lead sulfate is dried, and its mass is measured to be 0.1786 g. What was the concentration of sulfate in the original wastewater sample? SO_4^2- (aq) + Pb(O_3)_2 (aq) rightarrow PbSO_4(s) + 2NO_3^-(aq) 0.005889 M 5.417 M 169.8 M 16.98 M 0.5417 M In a volumetric analysis experiment, a solution of sodium oxalate (Na_2C_2O_4) in acidic solution t titrated with a solution of potassium permanganate (KMnO_4) according to the following balance chemical equation: 2KMnO_4(aq) + 8H_2SO_4(aq) + 5Na_2C_2O_4(aq) rightarrow 2MnSO_4(aq) + 8H_2O(l) + 10CO_2(g) + 5Na_2SO_4(aq) + K_2SO_4(aq) What volume of 0.0377 M KMnO_4 is required to titrate 0.132 g of Na_2C_2O_4 dissolved in 35.0 of solution? 1.40 mL 10.5 mL 26.1 mL 3.50 mL 35.0 mL

Explanation / Answer

molar mass of NaI= 149, moles = mass/molar mass = 6.9/149 =0.046

molarity= moles/L= 0.046*1000/38 M=1.21 M ( E is correct

2. molar mass of NaOH= 40 molarity = 27/(40*1.1) = 0.613 M ( E is correct)

3. moels of lead sulfate = 0.1786/303.3 = 0.000589 moels of SO4-2 is also = 0.000589

concentration = 0.000589/0.1 L= 0.00589 M ( A is correct)

4. 5 moles of Na2C2O4 requires 2 moles of KMnO4

moles of Na2C2O4 in 0.132 gm =0.132/134 =0.000985

moles of KMNo4= 0.000985*2/5= 0.000394

volume = moles/molarity = 0.000394/0.03777=0.010L = 10.43 ml ( b is correct)

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