Chemistry Molarity Quiz

Molarity is the number of moles of solute present in one liter of solution. It is calculated using the formula $M = \frac{moles}{liters}$.

The relationship is given by $moles = \frac{g}{form.wt}$, meaning moles are calculated by dividing the mass of the substance by its formula weight.

The mmoles of a titrant can be calculated using the formula $mmoles = (M)(ml)$ where M is the molarity and ml is the volume in milliliters.

<p>The reacting ratio is significant because it indicates the proportions of moles of titrant and titrated substance in the reaction, derived from the balanced equation.</p> Signup and view all the answers

<p>The formula is $g = (liters)(M)(form.wt)$, where liters is the volume in liters, M is the molarity, and form.wt is the formula weight.</p> Signup and view all the answers

<p>Molarity is calculated using the formula $M = \frac{0.002 \text{ mol}}{0.0367 \text{ L}}$, resulting in 0.0548 M.</p> Signup and view all the answers

<p>Normality is preferred because it accounts for reacting ratios that are not 1:1, allowing for more accurate calculations.</p> Signup and view all the answers

<p>An equivalent signifies the amount of a substance that can react with one Avogadro number of hydrogen ions (H+).</p> Signup and view all the answers

<p>1 equivalent of a substance corresponds to the grams that can react with one mole of H+, such as 36.5 g of HCl being 1 equivalent.</p> Signup and view all the answers

<p>The moles of KHP are calculated as $\frac{410.4 \text{ mg}}{204.2 \text{ g/mol}} = 0.00200 \text{ mol}$.</p> Signup and view all the answers

<p>Normality is defined as the number of equivalents of solute per liter of solution, which relates to equivalent weight by the equation: $eq.wt = \frac{form.wt}{No.ofH^+}$.</p> Signup and view all the answers

<p>The normality of the HCl solution is calculated using the equation: $N_{HCl} = \frac{(32.20)(0.0950)}{25.00}$ resulting in $N_{HCl} = 0.1224$.</p> Signup and view all the answers

<p>At the endpoint of a titration, the milliequivalents of titrant A equals the milliequivalents of titrated substance B, ensuring stoichiometric balance.</p> Signup and view all the answers

<p>Normality and molarity are directly related in titration calculations through the equation: $m_1A N_A = m_2B N_B$.</p> Signup and view all the answers

<p>The equivalent weight for H2SO4 in this scenario is determined using the formula: $eq.wt = \frac{form.wt}{2}$, since it can donate two protons.</p> Signup and view all the answers

<p>The moles of substance B are determined using the combining ratio (R), based on the stoichiometric coefficients of B and A in the balanced equation.</p> Signup and view all the answers

<p>The weight of substance B in milligrams is calculated using the formula mgB = mla * (MA / R) * (form. wt. B).</p> Signup and view all the answers

<p>The percentage of substance B is calculated by dividing the weight of B in milligrams by the weight of the sample in milligrams and multiplying by 100.</p> Signup and view all the answers

<p>The combining ratio (R) represents the ratio of the stoichiometric coefficients of substances in the balanced chemical equation, essential for mole calculations.</p> Signup and view all the answers

<p>The weight of acetic acid present is 210.1 mg CH₃COOH.</p> Signup and view all the answers

<p>The molarity of a solution can be determined by titrating a primary standard with known weight and calculating using the formula (mla)(MA)(R)(form wt. B) = mgB.</p> Signup and view all the answers

<p>The calculation of weight of substance B depends on the millimoles of A, its molecular weight, and the combining ratio (R).</p> Signup and view all the answers

<p>The reaction ratio (R) serves as a multiplier, adjusting the moles of substance A to determine the proportional moles of substance B produced.</p> Signup and view all the answers