A mixture contains 5% NaCl particles and 95% NaBr particles.If 10³ particles are taken,what is the relative standard deviation for NaCl and NaBr? A)13.78% for NaCl 0.73% for NaBr B)186.2% for NaCl 9.80% for NaBr C)95.00% for NaCl 5.00% for NaBr D)54.39% for NaCl 2.86% for NaBr E)6)89% for NaCl 0.36% for NaBr

Question

Quizplus · Accepted Answer

The relative standard deviation (RSD) is calculated using the formula RSD = (standard deviation / mean) * 100%. For NaCl, the mean is 50 particles (5% of 1000), and the standard deviation is the square root of 50, which gives an RSD of approximately 13.78%. For NaBr, the mean is 950 particles (95% of 1000), and the standard deviation is the square root of 950, which gives an RSD of approximately 0.73%.

A Mixture Contains 5% NaCl Particles and 95% NaBr Particles