You have 0.2231 grams of an unknown monovalent salt (MCl). You treat it with an excess of AgNO3 and precipitate all the chloride ion. 0.5471 grams of AgCl precipitates. What is the formula weight and identity of M

Formula weight of MCl = 58.55 g/mol

M = Sodium

Explanation:

Calculation of the moles of as:-

Mass = 0.5471 g

Molar mass of = 143.32 g/mol

The formula for the calculation of moles is shown below:

Thus,

1 mole of AgCl is produced from 1 mole of MCl

So, moles of MCl = 0.00381 mol

Mass = 0.2231 grams

Molar mass = Mass/Moles = 0.2231 grams/0.00381 mol= 58.55 g/mol

Molar mass of Cl = 35.5 g/mol

Molar mass of M = Molar mass of MCl - Molar mass of Cl = 58.55 g/mol - 35.5 g/mol = 23.05 g/mol

This mass corresponds to sodium.

answer

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hope it

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