Home
Class 12
CHEMISTRY
KCl and NaCl have fcc lattice. Calculate...

KCl and NaCl have fcc lattice. Calculate the ratio of density of NaCl to that of KCl if the ratio of edge of NaCl to that of KCl is 0.875.

Text Solution

AI Generated Solution

The correct Answer is:
To calculate the ratio of the density of NaCl to that of KCl given that the ratio of the edge lengths of NaCl to KCl is 0.875, we will follow these steps: ### Step 1: Understand the formula for density The density (D) of a crystal can be calculated using the formula: \[ D = \frac{Z \times M}{A^3 \times N_A} \] where: - \( Z \) = number of formula units per unit cell (for FCC, \( Z = 4 \)) - \( M \) = molar mass of the compound - \( A \) = edge length of the unit cell - \( N_A \) = Avogadro's number (constant, cancels out when taking ratios) ### Step 2: Calculate the molar masses of NaCl and KCl - Molar mass of NaCl: - Na = 23 g/mol - Cl = 35.5 g/mol - \( M_{NaCl} = 23 + 35.5 = 58.5 \, \text{g/mol} \) - Molar mass of KCl: - K = 39 g/mol - Cl = 35.5 g/mol - \( M_{KCl} = 39 + 35.5 = 74.5 \, \text{g/mol} \) ### Step 3: Set up the density ratio Using the density formula, we can express the densities for NaCl and KCl: - Density of NaCl: \[ D_{NaCl} = \frac{4 \times 58.5}{A_{NaCl}^3 \times N_A} \] - Density of KCl: \[ D_{KCl} = \frac{4 \times 74.5}{A_{KCl}^3 \times N_A} \] ### Step 4: Calculate the ratio of densities Now we can calculate the ratio of the densities: \[ \frac{D_{NaCl}}{D_{KCl}} = \frac{\frac{4 \times 58.5}{A_{NaCl}^3}}{\frac{4 \times 74.5}{A_{KCl}^3}} = \frac{58.5 \times A_{KCl}^3}{74.5 \times A_{NaCl}^3} \] ### Step 5: Substitute the ratio of edge lengths Given that the ratio of edge lengths \( \frac{A_{NaCl}}{A_{KCl}} = 0.875 \), we can express \( A_{NaCl} \) in terms of \( A_{KCl} \): \[ A_{NaCl} = 0.875 \times A_{KCl} \] Thus, \[ A_{NaCl}^3 = (0.875 \times A_{KCl})^3 = 0.875^3 \times A_{KCl}^3 \] ### Step 6: Substitute back into the density ratio Substituting \( A_{NaCl}^3 \) into the density ratio: \[ \frac{D_{NaCl}}{D_{KCl}} = \frac{58.5 \times A_{KCl}^3}{74.5 \times (0.875^3 \times A_{KCl}^3)} = \frac{58.5}{74.5 \times 0.875^3} \] ### Step 7: Calculate \( 0.875^3 \) Calculating \( 0.875^3 \): \[ 0.875^3 = 0.669921875 \] ### Step 8: Final calculation of the density ratio Now substituting this value back: \[ \frac{D_{NaCl}}{D_{KCl}} = \frac{58.5}{74.5 \times 0.669921875} \approx \frac{58.5}{49.89} \approx 1.172 \] ### Conclusion Thus, the ratio of the density of NaCl to that of KCl is approximately: \[ \frac{D_{NaCl}}{D_{KCl}} \approx 1.172 \]

To calculate the ratio of the density of NaCl to that of KCl given that the ratio of the edge lengths of NaCl to KCl is 0.875, we will follow these steps: ### Step 1: Understand the formula for density The density (D) of a crystal can be calculated using the formula: \[ D = \frac{Z \times M}{A^3 \times N_A} \] where: - \( Z \) = number of formula units per unit cell (for FCC, \( Z = 4 \)) - \( M \) = molar mass of the compound ...
Promotional Banner

Topper's Solved these Questions

  • SOLID STATE

    MODERN PUBLICATION|Exercise REVISION EXERCISES (NUMERICAL PROBLEMS)(CBSE Qs)|6 Videos
  • SOLID STATE

    MODERN PUBLICATION|Exercise HOTS - HIGHER ORDER THINKING SKILLS (ADVANCED LEVEL)|12 Videos
  • SOLID STATE

    MODERN PUBLICATION|Exercise REVISION EXERCISES (LONG ANSWER QUESTIONS)|6 Videos
  • POLYMERS

    MODERN PUBLICATION|Exercise Competition file (OBJECTIVE TYPE QUESTIONS) (C. MULTIPLE CHOICE QUESTIONS)(Integer Type Questions)|6 Videos
  • SOLUTIONS

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|14 Videos

Similar Questions

Explore conceptually related problems

KCI crystallizes in the same type of lacttice as does NaCl . Given that (r_(Na^(o+)))/(r_(Cl^(Θ))) = 0.5 and (r_(Na^(o+)))/(r_(K^(o+))) = 0.7 Calculate (a) the ratio of side of the unit cell for KCl to that for NaCl , and (b) the ratio of density of NaCl to that KCl .

KCl has NaCl type face centred cubic crystal structure and CsF has CsCl type cubic crystal structure. Calculate the ratio of densities of CsF and KCl it is given that the molar mass of CsF is two that of KCl and edge length of KCl unit cell of 1.5 times that for CsF .

CsF crystallizes in CsCl type structure and "KCl" in "NaCl" type structure. Edge-length of unit cell of KCl" is twice that of "CsF" and molecular weight of "CsF" is twice that of "KCl" .Calculate ratio of density of "CsF" to that of "KCl".

(a) NaCl has fcc structure. Calculate the number of NaCl units in a unit cell of NaCl. (b) Calculate the density of NaCl if edge length of NaCl unit cell is 564 pm. (Molar mass of NaCl = 58.5 g mol^(-1))

The lattice energies of NaCl, NaF, KCl and RbCl follow the order