What is the total molar concentration of ions if we mix together100. mL of 2.0 M

Question

What is the total molar concentration of ions if we mix together100. mL of 2.0 M NaNO_3 with 200. mL of 2.0 M KC_2H_3O_2. Assume the volumes are additive and both salts are fully dissociated 4.0 M 5.0 M 1.0 M 3.0 M 2.0 M Calculate the total molar concentration of ions when 30.0 g of KNO_3 (molar mass = 101.1 g/mol) is dissolved in enough water to make solution 1.98 M 5.98 M 3.98 M 2.98 M 4.98 M 14. What is the boiling point of solution of CCl_4 prepared by dissolving 12.0 g of I_2 (molar mass = 253.8 g/mol) in 200.0 g of CCl_4 (K_b of CCl_4=5.03 degree C/m. Boiling point of CCL_4=76.72 degree C) 77.56 degree C 76.72 degree C 76.91 degree C 77.91 degree C 78.12 degree C

Explanation / Answer

Moles of NaNO3 in 100ml of 2M= Molarity* Volume(L)= 2*100/1000= 0.2 moles

According to the reaction, NaNO3 -------->Na+ + NO3

0.2 mole of NaNO3 when fully ionized gives 0.2 mole of Na+ and 0,2 mole of NO3-

Moles of KC2H3O2 in 200ml of 2M= 2*200/1000=0.4 moles

KC2H3O2------------>K++ C2H3O2-, 0,4 mole of KC2H3O2 gives 0.4 mole of K+ and 0.4 moles of C2H3O2-.

hence total moles of ions = 0.2+0.2+0.4+0.4= 1.2

Totla moles= 0.2+0.4= 0.6, total volume= 100+200 =300ml =300/1000 =0.3L

Total concentration = moles/ Volume =1.2/0.3= 4M

This is the total concentration of ions in solution ( A is correct)

2.

Moles = mass/molar mass = 30/101.1 =0.296 moles

KNO3 dissociates as KNO3-->K+ +NO3-, 0.296 moles of KNO3 gives 0.296 moles of K+ and 0.296 moles of NO3-. Hence total moles of ions = 0.296+0.296=0.592 moles

Concentration of total ions= moles/ Volume(L)= 0.592/(300/1000)= 1.98M ( A is correct)

3. moles of I2= mass/molar mass = 12/253.8=0.0473, mass of solvent = 200 gm =200/1000 kg =0.2 Kg

molality, m= moles of solute/ kg of solvent CCl4 =0.0473/0.2 kg =0.2365,

Boiling point elevarion = i*kb*m, i= Van't Hoff factor for CCl4 =1

hence boiling point elevation =1*5.03*0.2365=1.189

boiling point = boiling point of solvent +1.189= 76.72+1.189=77.909 deg.c( d is correct)

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