What is the molarity of 350 mL solution that has had 32.0 g of NaCl dissolved in it?

Molarity = 1.57 M

Explanation:

Given data:

Volume of solution = 350 mL (350 mL × 1L /1000 ml = 0.35 L)

Mass of NaCl = 32.0 g

Molarity of solution = ?

Solution:

Number of moles of solute:

Number of moles = mass/molar mas

Number of moles = 32.0 g/ 58.44 g/mol

Number of moles = 0.55 mol

Molarity:

Molarity = number of moles / volume in L

Molarity = 0.55 mol / 0.35 mL

Molarity = 1.57 mol/L

Molarity = 1.57 M

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

plasma - state of matter made up of electrically charged particles