Question

calculate M of 30 g of Co(NO3)2.6H20 in 4.3L of solution

Answer 1

ans 0.02 M
30/3456= 0.08 mole , 0.08/4.3= 0.02 M

Answer 2

$\mathcal{ \huge \: \underline{ \underline{Question }}\: : }$

$\\ \sf Calculate~the ~ molarity~ of~ 30g~of~Co(NO3)2.6H2O~in~4.3L~of~soln.$

$\\ \bf{ \huge \: \underline{ \underline{Answer}}\: : }$

$\\ \tt \bold{ \underline{ given \:} : }$

$\\ \tt \: mass \: \: of \: \: solute \: \: = \: 30g \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \tt \: volume \: \: of \: \: soln. \: = \: 4.3 \: litre \: \: \: \: \:$

$\\ {\boxed{ \bf \: molarity \: \: = \: \: \frac{moles \: \: of \: \: solute}{volume \: \: of \: \: soln. \: \: in \: litre} }}$

■ Moles of the solute is not given .

$\\ \boxed{ \sf \: moles \: \: of \: \: solute \: = \frac{mass \: \: of \: \: solute}{molar \: \: mass \: \: of \: \: solute} }$

■ Molar mass of the solute is not given .

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Molar mass of Co(NO3)2.6H2O :

• Molar mass of COBALT = 59 g/mol
• Molar mass of NITROGEN = 14 g/mol
• Molar mass of OXYGEN = 16 g/mol
• Molar mass of HYDROGEN = 1 g/mol

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59 + 2 ( 14 + 3 × 16 ) + 6 ( 2 × 1 + 16 )

= 291 g/mol

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$\\ \sf \: moles \: \: of \: \: solute \: \: = \: \frac{30}{291} mol$

$\\ = \: \boxed{ \tt0.103 \: \: mol }$

$\bf \: molarity \: \: = \: \: \frac{0.103 \: \: mol}{4.3 \: \: litre} \\ \\ \bf \: = \underline{\underline{0.023 \: M}}$

$\\ \\ \bf \: Hope~~this~~helps~~you~!$