Consider mixing an excess of lead(II) nitrate (aq) with 200.0 mL of 0.400 M sodium chloride....

Consider mixing an excess of lead(II) nitrate (aq) with 200.0 mL of 0.400 M sodium chloride. Determine the mass of solid formed, assuming a complete reaction

What volume (in mL) of 0.800 M lead(II) nitrate must you add to make sure you make the mass of product you calculated in problem 3?

A 0.1044 g sample of the salt MCl was dissolved in water (making it in the aqueous phase). An excess of AgNO3 (aq) was added, precipitating 0.0889 g AgCl. What is the identity of M?

Expert Solution

Now given that 0.1044g of MCl solution added with AgNO3 to give 0.0889 g AgCl.

Hence reaction can be written as

MCl + AgNO3 —> AgCl + MNO3

So 1mole of MCl react with 1mole of AgNO3 to produce 1mole of AgCl.

As 0.0889g AgCl produce so Mole of AgCl = 0.0889/143.32 = 0.00062mole

Hence mole of MCl is also 0.00062mole

So molar mass of MCl is 0.1044/0.00062 = 168.32g

Molar mass of Cl is 35g

So molar mass of M is 168.32-35 = 133.32g

Which is approximately equal to molar mass of Cesium.

Hence, M is Cesium.

queen_honey_blossom answered 8 months ago

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