Question

wthat is the mass of the precipitate formed when 50ml of 16•9%solution of AgN03 is mixed with 50ml of 5•8NaCL solution ​

Answer 1

Answer:

$\huge{\underline{\overline{\color{darkblue}{\bf{\mid{♡AngelicDoll♡}\mid}}}}}$

Moles of AgNO3 = ${50\times16\times\frac{9}{100}\times169.8}$

→ ${0.05} {mol}$

Moles of NaCl = ${50\times5\times\frac{8}{100}\times58.5}$

→ ${0.05} {mol}$

$\bf\underline\orange{Reaction~as~follows}$

AgNO3 + NaCl => AgCl + NaNO3

Mass of AgCl = Mole of reaction × molar mass

→ ${0.05}\times{143.5}$

→ $\huge\bf\pink{7.16}{g}$

$<marquee>$

$\huge\red{AngelicDoll..!!!}$