Question

calculate the number of hydroxide ions present in 1.75 g of ammonium hydroxide

Answer 1

Number of hydroxide ions in 1.75 of ammonium hydroxide is 3.01*0^22 ions.

We know formula of ammonium hydroxide is NH4OH

So mass of ammonium hydroxide is 14+16+5 = 35 g.

Given mass of ammonium hydroxide is 1.75 g.

So moles of ammonium hydroxide is = (given weight/molecular weight)

Moles of NH4OH = 1.75/35 = 0.05 moles of ammonium hydroxide.

According to equation we have:-

NH4OH------>NH4^+1 + OH^-1

So 0.05 moles of ammonium hydroxide give 0.05 moles of hydroxide ions.

According to law we know any substance 1 mole has 6.022*10^23 molecules.

So 1 mole of  hydroxide has 6.022*10^23 ions of hydroxide.

So 0.05 moles of hydroxide has =0.05*6.022*10^23 ions of hydroxide.

= 3.011*10^22 ions of hydroxide.

So we have 1.75 g of ammonium hydroxide(NH4OH) has 3.011*10^22 ions of hydroxide(OH^-)