1. 75.0 mL of 1.45 M NaOH is neutralized by 93.7 mL of an HCl solution. What is

Question

1. 75.0 mL of 1.45 M NaOH is neutralized by 93.7 mL of an HCl solution. What is the molarity of the HCl solution? (Show your work.)

2. A sample of helium gas occupies 1,021 mL at 679 mmHg. For a gas sample at a constant temperature, determine the volume of helium at 745 mmHg. Show your work.

3. A gas at a temperature of 195 degrees C occupies a volume of 165 mL. Assuming constant pressure, determine the volume at 45 degrees C. Show your work

4. Calculate the volume, in liters, of a 4.75 mol H2S(g) at 90 degrees C and 1.75 atm. Show your work

Explanation / Answer

Answer - 1) We are given, volume of NaOH = 75.0 mL, molarity of NaOH = 1.45 M

Volume of HCl = 93.7 mL , molarity of HCl = ?

We know the neutralization reaction

NaOH + HCl -------> NaCl + H2O

We need to calculate the moles of NaOH

We know,

Moles of NaOH = 1.45 M * 0.075 L

                           = 0.109 moles

From the balanced reaction

1 moles of NaOH = 1 moles of HCl

So, 0.109 moles of NaOH = ?

= 0.109 moles of HCl

Now we also given,

Molarity of HCl = 0.109 moles / 0.0937 L

                           = 1.16 M

2) we are given, V1 = 1021 mL, P1 =679 mmHg , P2 = 745 mmHg

We know Boyles law

P1V1 = P2V2

So,

V2 = P1V1 / P2

      = 1021 mL * 679 mmHg / 745 mmHg

    = 930.5 mL

3) T1 = 195 oC + 273 K = 468 K , T2 = 45 +273 = 318 K , V1 = 165 mL , V2 = ?

We know Charles law

V1/T1 = V2/T2

So, V2 = V1*T2/T1

            = 165 mL * 318 K / 468 K

            = 112.1 mL

4) we are given, moles, n = 4.75 moles , T = 90 +273 =363 K , P = 1.75 atm

We know Ideal gas law

PV=nRT

We know,

V = nRT/V

    = 4.75 moles * 0.0821 L.atm.mol-1.K-1 *363 K / 1.75 atm

    = 80.9 L

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