Question

please Answer these questions......​

Answer 1

Step-by-step explanation:

$\sf \star \: Question \: 5 \: \star \\ \\ \sf Given\:weight\:of\:{Na}_{2}C{O}_{3} \:(W)=1.06 \: gm \\ \sf Molecular\:weight \:of \:{Na}_{2}C{O}_{3} = 106 \: gm\:{mol}^{-1} \\ \sf Volume = 250 \:mL = 0.25 \: L \\ \\ \sf We \:know\:that, \\ \\ \sf Molarity = \frac{No\:of\:moles}{Volume} \\ \\ \sf \longrightarrow = \frac{W}{M × V} \\ \\ \sf = \frac{1.06}{106×0.25} \\ \\ \sf = {1}{25} \\ \\ \sf = 0.04 M \\ \\ \sf so\:the\: molarity\:of\:the\:solution \: is 0.04 M$

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Answer 2

Your answer⬇️

$molecular \: weight \: of \: Na2Co3 = 106gm \: mol$

$given \: weight \: of \: Na2Co3(w) = 1.06gm$

$volume = 250ml \:= 0.25L$

$we \: know \: \: \: \: molarity = \frac{no. \: of \: moles}{volume}$

$= \frac{w}{m \times v}$

$= \frac{1.06}{106 \times 0.25}$

$= 125$

$= 0.04m$

I hope this will help you dude