Question

${\huge{\mathscr{AnsweR~IT}}}$
Q in Attachment !
# Chemistry 11th ​

Answer 1

${\huge{\boxed{\bf{Answer}}}}$

$\pink \star \: \: \mathrm \blue{300ppm \: of \: CaCO_{3} }$

$\star \: \: \: \mathrm {Refer \: to \: the \: attachment}$

Answer 2

$\huge\mathcal{\boxed{\underline{\color{olive} Good\: Evening}}}$

$\huge\mathcal{\underline{\color{aqua}QUESTION}}$

✨ 100 ml of tap water containing $\bf{Ca(HCO_3)_2}$ was titrated with N/50 HCl with methyl orange as indicator. If 30ml of HCl were required, calculate the temporary hardness as part of $\bf{CaCO_3}$ per 10⁶ parts of water.

$\huge{\orange{\boxed{\fcolorbox{lime}{cadetblue}{\pink{ANSWER}}}}}$

$\red\bullet\:\bf{M_{eq.}~of~CaCO_3~=~M_{eq}~of~Ca(HCO_3)_2~=~M_{eq}~of~HCl~}$

Where,

• $\bf{M_{eq}~=~Milli~equivalent~}$

We know that,

$\checkmark\:\sf\pink{N_1~V_1~=~N_2~V_2~}$

$\bf{\implies\:1000\times{\dfrac{w}{100/2}}~=~\dfrac{30}{50}\:}$

$\bf\green{\implies\:w~=~0.03~g~}$

So,

→ For 100 ml the amount of $\bf{CaCO_3}$ = 0.03 g

→ For 10⁶ ml = $\sf{\dfrac{0.03}{100}\times{10^6}}$

→ For 10⁶ ml = 300 ppm