Question
How many $\mathrm{mEq}$ of $\mathrm{Na}^{+}$are in $120 \mathrm{~mL}$ of an $8 \%$ solution of sodium cyclamate $($ molecular weight $=201.33 \mathrm{~g} / \mathrm{mol} ;$ valence $=1)$ ?
Step 1
Given that the solution is $8\%$ sodium cyclamate, we can calculate the amount of sodium cyclamate in $120 \mathrm{~mL}$ of the solution. $8\% = \frac{8}{100} = 0.08$ Amount of sodium cyclamate $= 0.08 \times 120 \mathrm{~mL} = 9.6 \mathrm{~mL}$ Show more…
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Topic 2 : Stoichiometry, Equivalent Concept, Neutralization and Redox Titration
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