Calculate the mass in gram of binormal solution in NAOH

80 grams

Explanation:

Normality is one of the units of concentration and it is derived by dividing the gram or mole equivalent of the solute in a solution by the volume (in liters) of the solution.  

The equivalent weight of NaOH is derived by dividing the molecular weight of the compound by 1. Hence, the equivalent weight of NaOH in 1 normal solution is 40/1 = 40 g.

If 1 normal NaOH solution contains 40 g of NaOH in 1 liter of the solution, then a binormal (2 normal) solution of NaOH would contain;

   2 x 40 = 80 g of NaOH.

hey there!

boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. the vapor pressure of solvent molecules is decreased when a solute is added, so a higher temperature is required to increase the number of solvent molecules in the gas phase above the liquid.

at the freezing point, the vapor pressures of the solid and liquid are equal; a lower temperature is needed to reduce the number of solvent particles above the liquid.

hope this you

tobey

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol